A) \[nB\]
B) \[{{n}^{2}}B\]
C) \[2nB\]
D) \[2{{n}^{2}}B\]
Correct Answer: B
Solution :
(b) \[{{n}^{2}}B\]. Magnetic field at the centre of this wire. \[B=\frac{{{\mu }_{0}}l}{2R}\] (1) The same wire is bent into a circular loop of 'n' turns. New radius is given by. \[n\times 2\pi r=2\pi R\Rightarrow r=\frac{R}{n}\] Thus. magnetic field at the centre of this loop \[=\frac{{{\mu }_{0}}nl}{2r}\] \[=\frac{{{\mu }_{0}}nl}{2r}\times n={{n}^{2}}\frac{{{\mu }_{0}}l}{2R}={{n}^{2}}B\] (from(1))You need to login to perform this action.
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